Question

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Answer 1

f:R→R, x ·f(x) -(1-x)·f(-x) = x+1 unde x ∈R .

x = 1 => f(1) = 2

x = -1 => (-1)·f(-1) -2·f(1) = 0 dar deoarece f(1) = 2 => -f(-1) = 4 => f(-1) = -4 .

f(x) = a·x² +b·x +c unde a, b si c reale iar a ≠ 0 respectiv b ≠ 0 .

x·(a·x² +b·x +c) -(1-x)·(a·x² -b·x +c) = x+1 <=>

a·x³ +b·x² +c·x -a·x² +b·x -c +a·x³ +b·x² +c·x = x+1 <=>

a(2·x³ -x²) +b(2·x² +x) +c(2·x -1) = x+1 <=>

a=2, b=1 si c=-1 verifica ecuatia data deci functiile de gradul doi care verifica relatia data sunt de fotma f(x) = 2·x² +x -1 unde x ∈R .